Physical Chemistry Third Edition

208 5 Phase Equilibrium

The Clapeyron Equation

In a one-component system the chemical potential is equal to the molar Gibbs energy,

so that if phases I and II of a single substance are at equilibrium

G(I)mG(II)m (5.3-2)

We impose an infinitesimal changedTin the temperature of the system, maintaining

equilibrium during the change. SincePis a function ofT, the pressure will change

by an amountdPthat is determined bydT, and the molar Gibbs energies of the two

phases,G(I)mandG(II)m, will undergo changes that are given in terms ofdPanddTby

Eq. (4.2-19):

dG(I)m−Sm(I)dT+Vm(I)dP (5.3-3)

dG(II)m −Sm(II)dT+Vm(II)dP (5.3-4)

The molar Gibbs energies remain equal to each other after the change, so thatdG

(I)

m

dG(II)m, and

−Sm(I)dT+Vm(I)dP−Sm(II)dT+Vm(II)dP (5.3-5)

Nonrigorously “dividing” this equation bydT, we obtain theClapeyron equation:

dP

dT



Sm(II)−Sm(I)

V

(II)

m −V

(I)

m



∆Sm

∆Vm

(Clapeyron equation) (5.3-6)

The Clapeyronequationisnamed
after Benoit-Pierre-Emile Clapeyron,
1799–1864,aFrench engineer who
translatedCarnot’s cycle into the
language of calculus.

For a reversible phase change at constant pressure∆G0, so that

∆Sm

∆Hm

T

(5.3-7)

The Clapeyron equation can be written

dP

dT



∆Hm

T∆Vm

(another version of the

Clapeyron equation) (5.3-8)

We can use the Clapeyron equation to interpret the slopes of the curves in a phase

diagram.

EXAMPLE 5.2

Interpret the curves in the phase diagram of water, Figure 5.3, that appear to be vertical and

horizontal line segments.

Solution

A horizontal line segment corresponds to zero value ofdP/dT, implying that∆Hm0 and

∆Sm0 for the phase transition. For example, between ice VI and ice VII it appears that

∆Sm0 and∆Vm0. A vertical line segment corresponds to an undefined (infinite) value

fordP/dT, implying that∆Vm0 and∆Sm0. For example, ice VII and ice VIII appear

to have the same molar volume.